The figure here shows a plot of potential energy U versus position x of a 0.878 kg particle that can travel only along an x axis. (Nonconservative forces are not involved.) Three values are UA-15.0J, Us-35.0J and Uc-45.0 J. The particle is released at x -4.50 m with an initial speed of 7.99 m/s, headed in the negative x direction. (a) If the particle can reach x - 1.00 m, what is its speed there, and if it cannot, what is its turning point? What are the (b) magnitude and (c) direction of the force on the particle as it begins to move to the left of x - 4.00 m? Suppose, instead, the particle is headed in the positive x direction when it is released at x - 4.50 mat speed 7.99 m/s. (d) If the particle can reach x - 7.00 m, what is its speed there, and if it cannot, what is its turning point? What are the (e) magnitude and (f) direction of the force on the particle as it begins to move to the right of x - 5.00 m? (0)2 Ucr UB S 4 x (m) 1 6

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The figure here shows a plot of potential energy U versus position x of a 0.878 kg particle that can travel only along an x axis.
(Nonconservative forces are not involved.) Three values are UA-15.0J, Us - 35.0 J and Uc-45.0 J. The particle is released at x
-4.50 m with an initial speed of 7.99 m/s, headed in the negative x direction. (a) If the particle can reach x -1.00 m, what is its speed
there, and if it cannot, what is its turning point? What are the (b) magnitude and (c) direction of the force on the particle as it begins to
move to the left of x- 4.00 m? Suppose, instead, the particle is headed in the positive x direction when it is released at x - 4.50 m at
speed 7.99 m/s. (d) If the particle can reach x - 7.00 m, what is its speed there, and if it cannot, what is its turning point? What are the
(e) magnitude and (f) direction of the force on the particle as it begins to move to the right of x-5.00 m?
(0)2
Ucr
UB
UA
2
(a) Number
(b) Number 10
4
x (m)
(c) positive (+x)
(e) Number
(f)
4.30
(d) Number i -43
-30
negative (-x) ✔
6
Unit
Unit
Unit
m/s
N
m
Unit N
Transcribed Image Text:The figure here shows a plot of potential energy U versus position x of a 0.878 kg particle that can travel only along an x axis. (Nonconservative forces are not involved.) Three values are UA-15.0J, Us - 35.0 J and Uc-45.0 J. The particle is released at x -4.50 m with an initial speed of 7.99 m/s, headed in the negative x direction. (a) If the particle can reach x -1.00 m, what is its speed there, and if it cannot, what is its turning point? What are the (b) magnitude and (c) direction of the force on the particle as it begins to move to the left of x- 4.00 m? Suppose, instead, the particle is headed in the positive x direction when it is released at x - 4.50 m at speed 7.99 m/s. (d) If the particle can reach x - 7.00 m, what is its speed there, and if it cannot, what is its turning point? What are the (e) magnitude and (f) direction of the force on the particle as it begins to move to the right of x-5.00 m? (0)2 Ucr UB UA 2 (a) Number (b) Number 10 4 x (m) (c) positive (+x) (e) Number (f) 4.30 (d) Number i -43 -30 negative (-x) ✔ 6 Unit Unit Unit m/s N m Unit N
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